Leetcode_valid-palindrome

地址：http://oj.leetcode.com/problems/valid-palindrome/

Given a string, determine if it is a palindrome, considering only alphanumeric characters and ignoring cases.

For example,

"A man, a plan, a canal: Panama"

is a palindrome.

"race a car"

is not a palindrome.

Note:

Have you consider that the string might be empty? This is a good question to ask during an interview.

For the purpose of this problem, we define empty string as valid palindrome.
思路：只把大小写字母和数字留下，用stl反置一下再对比是否相同。
另外有个函数可以检测是否是字母和数字。不过在这题里不是很合适，因为有大小写的缘故。

int isalnum ( int c )

参考代码：

class Solution {
public:
bool isPalindrome(string s) {
if(s.empty())
return true;
string str;
for(int i = 0; i <s.length(); ++i)
{
if((s[i]>='a'&&s[i]<='z') || (s[i]>='0'&&s[i]<='9'))
str+=s[i];
else if(s[i]>='A'&&s[i]<='Z')
str+=s[i]-'A'+'a';
}
string rev_str = str;
reverse(rev_str.begin(), rev_str.end());
return rev_str == str;
}
};

//SECOND TRIALclass Solution {public:    bool isPalindrome(string s) {        if(s.empty())            return true;        int cnt = 0;        for(int i = 0; i<s.length(); ++i)        {            if((s[i]>='0' && s[i]<='9')||(s[i]>='a' &&s[i]<='z'))                s[cnt++] = s[i];            else if(s[i]>='A' && s[i]<='Z')                s[cnt++] = s[i] + 'a' - 'A';        }        s = s.substr(0, cnt);        string revs(s.rbegin(), s.rend());        return revs==s;    }};